Question

$\underset{0}{\overset{2\pi }{\int }}\frac{dx}{1+{\tan }^{4}x}$ is equal to

• A. $\pi$
• B. ${\pi }^2$
• C. $0$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\pi$

$I=\underset{0}{\overset{2\pi }{\int }}\frac{dx}{1+{\tan }^{4}x}I=2\underset{0}{\overset{\pi }{\int }}\frac{dx}{1+{\tan }^{4}x}I=4\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{1+{\tan }^{4}x}\cdots \left(i\right)\{\because \underset{0}{\overset{2a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{when}f\left(x\right)=f(2a-x)\}I=4\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{1+{\tan }^4(\frac{\pi }{2}-x)}\Rightarrow I=4\underset{0}{\overset{\pi /2}{\int }}\frac{dx}{1+{\cot }^{4}x}\Rightarrow I=4\underset{0}{\overset{\pi /2}{\int }}\frac{{\tan }^{4}x}{1+{\tan }^{4}x}dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, we get
$2I=4\underset{0}{\overset{\pi /2}{\int }}1\cdot dx=4\cdot \frac{\pi }{2}\Rightarrow I=\pi$